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For the sake of curiosity, does anyone know what $\sup\limits_{n \in \mathbb{N}} \sum\limits_{k=0}^n \cos(k)$ or $\sup\limits_{n \in \mathbb{N}} \sum\limits_{k=0}^n \sin(k)$ are? It is pretty easy to see that both partial sums are bounded, but I don't even have a clue as to what their respective supremums should be. Definitely something not too large.

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    $\begingroup$ The both have closed form. See this. $\endgroup$ – dezdichado May 8 '17 at 4:05
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    $\begingroup$ Do you mean $\cos(k)$ and $\sin(k)$? $\endgroup$ – Bungo May 8 '17 at 4:10
  • $\begingroup$ The question is about $\sum cos(n)$ not $\sum cos(nx)$. $\endgroup$ – Mike A. May 8 '17 at 4:13
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    $\begingroup$ ^You can simply substitute $x = 1$. $\endgroup$ – JimmyK4542 May 8 '17 at 4:14
  • $\begingroup$ (I think this will work out) To find $\sup \sum cos(n)$, use the closed form of the sums to solve $\sum \sin(n) = 0$ for $n$ and then apply the floor function to this $n$ to get the value at which the supremum for $\sum cos(n) $ occures $\endgroup$ – infinitylord May 8 '17 at 4:29
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As shown in the link that dezdichado gave in the comments, we can write

$\displaystyle\sum_{k = 0}^{n}\cos(k) = \sum_{k = 0}^{n}\dfrac{\cos(k)\sin (\tfrac{1}{2})}{\sin (\tfrac{1}{2})} = \sum_{k = 0}^{n}\dfrac{\sin(\tfrac{k+1}{2})-\sin(\tfrac{k-1}{2})}{2\sin (\tfrac{1}{2})}$

$= \dfrac{\sin(\tfrac{n+1}{2})-\sin (-\tfrac{1}{2})}{2\sin (\tfrac{1}{2})} = \dfrac{\sin(\tfrac{n+1}{2})+\sin (\tfrac{1}{2})}{2\sin (\tfrac{1}{2})}$

and

$\displaystyle\sum_{k = 0}^{n}\sin(k) = \sum_{k = 0}^{n}\dfrac{\sin(k)\sin (\tfrac{1}{2})}{\sin (\tfrac{1}{2})} = \sum_{k = 0}^{n}\dfrac{\cos(\tfrac{k-1}{2})-\sin(\tfrac{k+1}{2})}{2\sin (\tfrac{1}{2})}$

$= \dfrac{\cos(-\tfrac{1}{2})-\cos(\tfrac{n+1}{2})}{2\sin (\tfrac{1}{2})} = \dfrac{\cos(\tfrac{1}{2})-\cos(\tfrac{n+1}{2})}{2\sin (\tfrac{1}{2})}$.

You can prove that $\displaystyle\sup_{n \in \mathbb{N}}\sin(\tfrac{n+1}{2}) = 1$ and $\displaystyle\inf_{n \in \mathbb{N}}\cos(\tfrac{n+1}{2}) = -1$ using arguments similar to what is discussed here.

Combining those results gives $\displaystyle\sup_{n \in \mathbb{N}}\sum_{k = 0}^{n}\cos(k) = \dfrac{1}{2\sin(\tfrac{1}{2})}+\dfrac{1}{2}$ and $\displaystyle\sup_{n \in \mathbb{N}}\sum_{k = 0}^{n}\sin(k) = \dfrac{\cos(\tfrac{1}{2})+1}{2\sin (\tfrac{1}{2})}$.

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